Question: You have found the following ages (in years) of all 5 tigers at your local zoo: $ 2,\enspace 6,\enspace 8,\enspace 3,\enspace 13$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{2 + 6 + 8 + 3 + 13}{{5}} = {6.4\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-4.4$ years $19.36$ years $^2$ $6$ years $-0.4$ years $0.16$ years $^2$ $8$ years $1.6$ years $2.56$ years $^2$ $3$ years $-3.4$ years $11.56$ years $^2$ $13$ years $6.6$ years $43.56$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{19.36} + {0.16} + {2.56} + {11.56} + {43.56}} {{5}} $ $ {\sigma^2} = \dfrac{{77.2}}{{5}} = {15.44\text{ years}^2} $ The average tiger at the zoo is 6.4 years old. The population variance is 15.44 years $^2$.